dxdy=rdrdΘ
Why do I get otherwise..?
I'm not looking for a graphical approach, I'm more so wondering why my mathematical approach is wrong
>>8048938
for starters, it looked like you multiplied wrong the first term in the expansion
>>8048943
my bad, there's supposed to be a cosine in there too, but i miscopied it from my scrap paper
>>8048938
Where do you get those initial equalities for dx and dy?
Not saying it's wrong, I just can't remember where I've seen that shown before
>>8048949
From the Total Derivative:
https://en.wikipedia.org/wiki/Total_derivative
Let [math]\omega = \operatorname{dx} \wedge \operatorname{dy} [/math]
and [math]f:{\mathbb{R}^2} \to {\mathbb{R}^2}[/math] such that[math]\left( {r,\theta } \right) \mapsto \left( {r\cos \theta ,r\sin \theta } \right) = \left( {x,y} \right)[/math] .
Then calculate the pullback...
[math]{f^*}\omega = {f^*}\operatorname{dx} \wedge {f^*}\operatorname{dy} = \operatorname{d} \left( {x \circ f} \right) \wedge \operatorname{d} \left( {y \circ f} \right) = \operatorname{d} \left( {r\cos \theta } \right) \wedge \operatorname{d} \left( {r\sin \theta } \right) = \left( {\cos \theta \operatorname{d} r - r\sin \theta \operatorname{d} \theta } \right) \wedge \left( {\sin \theta \operatorname{d} r + r\cos \theta \operatorname{d} \theta } \right) = r\operatorname{d} r \wedge \operatorname{d} \theta [/math]
wow so hard
>>8048955
Okay now explain it to me in layman's terms.
Does this mean you can't go about calculating dxdy the way i did and you HAVE to do it another way? such as the Jacobian? I'm not a mathematician wiz so I'm just looking for som clarification
>>8048964
Laymans terms: that's not how multiplication of differentials works. This was an annoyance of the notation in classical calculus, that you had to be careful about order/etc. to recover the proper area/volume/etc. element from a 'product' of the differentials. Also, differentials, even informally, were built so that their squared products vanished (i.e., you should only look at first order in a given differential).
The *proper* way to obtain the new area/volume/etc. element, even in classical calculus, is to use a jacobian transformation. This is the change of variables formula that correctly transforms your integration.
What the other user posted is a rigorous construction of forms that has the nice benefit of matching all the classical calculus formulae and intuition. The new rules are that product between differentials is an anti-commutating product (dx dy = -dy dx) and so automatically the square of a differential vanishes (i.e., dx dx = 0).
TLDR: straight multiplication of differentials has always been wrong, even when you assume a squared differential vanishes. You have to use the jacobian in classical calculus, or learn about differential forms and their algebra to do things properly.
>>8049030
Thanks! I never knew. Thanks for the eye opener! Best to you
>>8048938
>>8048964
You kinda did do it right, you just clearly weren't aware of the proper multiplication.
You multiply differentials using the wedge product [math]\wedge [/math].
Two properties this product has when acting on differentials / 1-forms is that:
[math] \operatorname{dx} \wedge \operatorname{dx} = 0 [/math]
i.e. any 1-form squared is zero
[math] \operatorname{dx} \wedge \operatorname{dy} = - \operatorname{dy} \wedge \operatorname{dx} [/math]
i.e. It is anti-commutative for 1-forms
If you take the above two properties into account, the expression you calculated simplifies to the correct answer.
